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МатематикаГДЗ по алгебре 7 класс Макарычев ФГОС Задание 540
Задание 540
\[\boxed{\text{540.}\text{\ }еуроки - ответы\ на\ пятёрку}\]
Пояснение.
Чтобы узнать, какие числа являются корнями, подставим их в уравнения и вычислим. Если равенство верное, то число – корень уравнения.
Решение.
\[\textbf{а)}\ x^{4} = 81\]
\[x = - 3:\]
\[( - 3)^{4} = 81\]
\[81 = 81.\]
\[x = - 2:\]
\[( - 2)^{4} = 81\]
\[16 \neq 81.\]
\[x = - 1:\]
\[( - 1)^{4} = 81\]
\[1 \neq 81.\]
\[x = 1:\]
\[1^{4} = 81\]
\[1 \neq 81.\]
\[x = 2:\]
\[2^{4} = 81\]
\[16 \neq 81.\]
\[x = 3:\]
\[3^{4} = 81\]
\[81 = 81.\]
\[Ответ:x = \pm 3.\]
\[\textbf{б)}\ x^{6} = 64\]
\[x = - 3:\]
\[( - 3)^{6} = 64\]
\[729 \neq 64.\]
\[x = - 2:\]
\[( - 2)^{6} = 64\]
\[64 = 64.\]
\[x = - 1:\]
\[( - 1)^{6} = 64\]
\[1 \neq 64.\]
\[x = 1:\]
\[1^{6} = 64\]
\[1 \neq 64.\]
\[x = 2:\]
\[2^{6} = 64\]
\[64 = 64.\]
\[x = 3:\]
\[3^{6} = 64\]
\[729 \neq 64.\]
\[Ответ:x = \pm 2.\]
\[\textbf{в)}\ x^{2} - x = 2\]
\[x = - 3:\]
\[( - 3)^{2} - ( - 3) = 2\]
\[9 + 3 = 2\]
\[11 \neq 2.\]
\[x = - 2:\]
\[( - 2)^{2} - ( - 2) = 2\]
\[4 + 2 = 2\]
\[6 \neq 2.\]
\[x = - 1:\]
\[( - 1)^{2} - ( - 1) = 2\]
\[1 + 1 = 2\]
\[2 = 2.\]
\[x = 1:\]
\[1^{2} - 1 = 2\]
\[1 - 1 = 2\]
\[0 \neq 2.\]
\[x = 2:\]
\[2^{2} - 2 = 2\]
\[4 - 2 = 2\]
\[2 = 2.\]
\[x = 3:\]
\[3^{2} - 3 = 2\]
\[9 - 3 = 2\]
\[6 \neq 2.\]
\[x = 2;\ \ \ x = - 1\]
\[Ответ:x = 2;\ \ x = - 1.\]
\[\textbf{г)}\ x^{4} + x^{3} = 6x^{2}\ \ \ |\ :x^{2}\]
\[x^{2} + x = 6\]
\[x = - 3:\]
\[( - 3)^{2} - 3 = 6\]
\[9 - 3 = 6\]
\[6 = 6.\]
\[x = - 2:\]
\[( - 2)^{2} - 2 = 6\]
\[4 - 2 = 6\]
\[2 \neq 6.\]
\[x = - 1:\]
\[( - 1)^{2} - 1 = 6\]
\[1 - 1 = 6\]
\[0 \neq 6.\]
\[x = 1:\]
\[1^{2} + 1 = 6\]
\[1 + 1 = 6\]
\[2 \neq 6.\]
\[x = 2:\]
\[2^{2} + 2 = 6\]
\[4 + 2 = 6\]
\[6 = 6.\]
\[x = 3:\]
\[3^{2} + 3 = 6\]
\[9 + 3 = 6\]
\[12 \neq 6.\]
\[x = - 3;\ \ x = 2\]
\[Ответ:x = - 3;x = 2.\]
\[\textbf{д)}\ x^{3} - 3x^{2} - 4x + 12 = 0\]
\[x = - 3:\]
\[( - 3)^{3} - 3 \cdot ( - 3)^{2} - 4 \cdot ( - 3) + 12 =\]
\[= 0\]
\[- 27 - 27 + 12 + 12 = 0\]
\[- 54 + 24 = 0\]
\[- 30 \neq 0.\]
\[x = - 2:\]
\[( - 2)^{3} - 3 \cdot ( - 2)^{2} - 4 \cdot ( - 2) + 12 =\]
\[= 0\]
\[- 8 - 12 + 8 + 12 = 0\]
\[0 = 0.\]
\[x = - 1:\]
\[( - 1)^{3} - 3 \cdot ( - 1)^{2} - 4 \cdot ( - 1) + 12 =\]
\[= 0\]
\[- 1 - 3 + 4 + 12 = 0\]
\[12 \neq 0.\]
\[x = 1:\]
\[1^{3} - 3 \cdot 1^{2} - 4 \cdot 1 + 12 = 0\]
\[1 - 3 - 4 + 12 = 0\]
\[6 \neq 0.\]
\[x = 2:\]
\[2^{3} - 3 \cdot 2^{2} - 4 \cdot 2 + 12 = 0\]
\[8 - 12 - 8 + 12 = 0\]
\[0 = 0.\]
\[x = 3:\]
\[3^{3} - 3 \cdot 3^{2} - 4 \cdot 3 + 12 = 0\]
\[27 - 27 - 12 + 12 = 0\]
\[0 = 0.\]
\[Ответ:x = \pm 2;\ \ x = 3.\]
\[\textbf{е)}\ x^{3} + 3x^{2} - x - 3 = 0\]
\[x = - 3:\]
\[( - 3)^{3} + 3 \cdot ( - 3)^{2} - ( - 3) - 3 =\]
\[= 0\]
\[- 27 + 27 + 3 - 3 = 0\]
\[0 = 0.\]
\[x = - 2:\]
\[( - 2)^{3} + 3 \cdot ( - 2)^{2} - ( - 2) - 3 =\]
\[= 0\]
\[- 8 + 12 + 2 - 3 = 0\]
\[3 \neq 0.\]
\[x = - 1:\]
\[( - 1)^{3} + 3 \cdot ( - 1)^{2} - ( - 1) - 3 =\]
\[= 0\]
\[- 1 + 3 + 1 - 3 = 0\]
\[0 = 0.\]
\[x = 1:\]
\[1^{3} + 3 \cdot 1^{2} - 1 - 3 = 0\]
\[1 + 3 - 1 - 3 = 0\]
\[0 = 0.\]
\[x = 2:\]
\[2^{3} + 3 \cdot 2^{2} - 2 - 3 = 0\]
\[8 + 12 - 2 - 3 = 0\]
\[15 \neq 0.\]
\[x = 3:\]
\[3^{3} + 3 \cdot 3^{2} - 3 - 3 = 0\]
\[27 + 27 - 3 - 3 = 0\]
\[48 \neq 0.\]
\[x = 1;\ \ x = - 1;\ \ x = - 3.\]
\[Ответ:x = \pm 1;\ \ x = - 3.\ \]
